Assignment 4/Chapter 5

Problem 3

We now review k-fold cross-validation.

(a) Explain how k-fold cross-validation is implemented.

• The purpose of cross-validation is that it estimates the test error rate by holding out a subset of the training observations from the fitting process and then applies the statistical learning method to those held out observations.

(b) What are the advantages and disadvantages of k-fold cross-validation relative to:

i. The validation set approach?

• The validation set approach is a simple strategy that involves randomly dividing the available set of observations into two parts, a training set and a validation set, otherwise known as the hold-out set. The largest advantage is that this strategy is conceptually simple and easy to implement, but has two potential drawbacks; First, the validation estimate of the test error can be highly variable depending on the data that was split into each half. Second, we run the risk of overestimating the test error rate because statistical methods tend to perform worse when trained on fewer observations and we have cut our data set in half for this test’s purpose.

ii. LOOCV?

• The Leave-one-out cross-validation is closely related to the above and attempts to address that method’s two drawbacks by creating a single observation used for the validation set and the remaining observations make up the training set.

• Advantages here are that LOOCV is far less bias due to refitting our training sets repeatedly, almost as many times as the entire data set contains. This helps in not overestimating. Additionally, there is no randomness in the training/validation set splits compared to the validation set approach.

• Disadvantages is that LOOCV can be expensive to implement as it takes much more time to fit n times and can be very time consuming.

iii.Comparison of K-fold CV vs The Validation Set Approach and LOOCV

• K-Fold has a huge computational advantage. Instead of fitting our data n times, we simply fit it K times for similar results and is less time consuming/expensive. K indicates how many times the data was split, but this variability is typically much lower than the variability in the test error estimates from the validation set approach.

Problem 5

In Chapter 4, we used logistic regression to predict the probability of default using income and balance on the Default data set. We will now estimate the test error of this logistic regression model using the validation set approach. Do not forget to set a random seed before beginning your analysis.

(a) Fit a logistic regression model that uses income and balance to predict default.

library(ISLR2)
attach(Default)
set.seed(1)
glm.fit=glm(default~income+balance,data=default,family=binomial)
summary(glm.fit)

## 
## Call:
## glm(formula = default ~ income + balance, family = binomial, 
##     data = default)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4725  -0.1444  -0.0574  -0.0211   3.7245  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.154e+01  4.348e-01 -26.545  < 2e-16 ***
## income       2.081e-05  4.985e-06   4.174 2.99e-05 ***
## balance      5.647e-03  2.274e-04  24.836  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 2920.6  on 9999  degrees of freedom
## Residual deviance: 1579.0  on 9997  degrees of freedom
## AIC: 1585
## 
## Number of Fisher Scoring iterations: 8

(b) Using the validation set approach, estimate the test error of this model. In order to do this, you must perform the following steps:

i. Split the sample set into a training set and a validation set.

train=sample(10000,5000)

ii. Fit a multiple logistic regression model using only the training observations.

glm.fit2=glm(default~income+balance,data=default,family=binomial,subset=train)
summary(glm.fit2)

## 
## Call:
## glm(formula = default ~ income + balance, family = binomial, 
##     data = default, subset = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.5830  -0.1428  -0.0573  -0.0213   3.3395  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.194e+01  6.178e-01 -19.333  < 2e-16 ***
## income       3.262e-05  7.024e-06   4.644 3.41e-06 ***
## balance      5.689e-03  3.158e-04  18.014  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1523.8  on 4999  degrees of freedom
## Residual deviance:  803.3  on 4997  degrees of freedom
## AIC: 809.3
## 
## Number of Fisher Scoring iterations: 8

iii. Obtain a prediction of default status for each individual in the validation set by computing the posterior probability of default for that individual, and classifying the individual to the default category if the posterior probability is greater than 0.5.

glm.pred=predict(glm.fit2,default[-train],type="response")
glm.predict=ifelse(glm.pred>0.5,"Yes","No")

iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.

mean(glm.predict != default[-train])

## [1] 0.0455

(c) Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.

glm.test={
train=sample(10000,5000)
glm.fit3=glm(default~income+balance,data=default,family=binomial,subset=train)
glm.pred3=predict(glm.fit3,default[-train],type="response")
glm.predict3=ifelse(glm.pred3>0.5,"Yes","No")}
glm.test2={
train=sample(10000,5000)
glm.fit4=glm(default~income+balance,data=default,family=binomial,subset=train)
glm.pred4=predict(glm.fit4,default[-train],type="response")
glm.predict4=ifelse(glm.pred4>0.5,"Yes","No")}
glm.test3={
train=sample(10000,5000)
glm.fit5=glm(default~income+balance,data=default,family=binomial,subset=train)
glm.pred5=predict(glm.fit5,default[-train],type="response")
glm.predict5=ifelse(glm.pred5>0.5,"Yes","No")}
mean(glm.predict3 != default[-train])

## [1] 0.0466

mean(glm.predict4 != default[-train])

## [1] 0.0464

mean(glm.predict5 != default[-train])

## [1] 0.0457

• After repeating the process three times, we have the test error rates that range from 4.57%-4.66% which shows that the rates vary depending on the training/validation set definitions.

(d) Now consider a logistic regression model that predicts the probability of default using income, balance, and a dummy variable for student. Estimate the test error for this model using the validation set approach. Comment on whether or not including a dummy variable for student leads to a reduction in the test error rate.

glm.student=glm(default~income+balance+student,data=default,family=binomial,subset=train)
glm.pred.student=predict(glm.student,default[-train],type="response")
glm.predict.student=ifelse(glm.pred.student>0.5,"Yes","No")
mean(glm.predict.student != default[-train])

## [1] 0.0455

• I wanted to follow the exact steps above when creating this new model to ensure we have an accurate comparison. When adding student into the logistic regression formula and completing a validation set test, we receive an error rate of 4.55% meaning we do technically see a decrease in error rate, but nothing significant or meaningful.

Problem 6

We continue to consider the use of a logistic regression model to predict the probability of default using income and balance on the Default data set. In particular, we will now compute estimates for the standard errors of the income and balance logistic regression coefficients in two different ways: (1) using the bootstrap, and (2) using the standard formula for computing the standard errors in the glm() function. Do not forget to set a random seed before beginning your analysis.

set.seed(100)

(a) Using the summary() and glm() functions, determine the estimated standard errors for the coefficients associated with income and balance in a multiple logistic regression model that uses both predictors.

glm.fit=glm(default~income+balance,data=Default,family=binomial,subset=train)
summary(glm.fit)

## 
## Call:
## glm(formula = default ~ income + balance, family = binomial, 
##     data = Default, subset = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4027  -0.1517  -0.0624  -0.0233   3.6833  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.112e+01  5.816e-01 -19.120   <2e-16 ***
## income       1.638e-05  6.755e-06   2.425   0.0153 *  
## balance      5.489e-03  3.067e-04  17.897   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1503.85  on 4999  degrees of freedom
## Residual deviance:  831.51  on 4997  degrees of freedom
## AIC: 837.51
## 
## Number of Fisher Scoring iterations: 8

(b) Write a function, boot.fn(), that takes as input the Default data set as well as an index of the observations, and that outputs the coefficient estimates for income and balance in the multiple logistic regression model.

boot.fn=function(data,index){
  boot=glm(default~income+balance,data=data,family=binomial,subset=index)
  return (coef(boot))
}

(c) Use the boot() function together with your boot.fn() function to estimate the standard errors of the logistic regression coefficients for income and balance.

library(boot)
boot(Default,boot.fn,1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01 -2.927307e-02 4.446538e-01
## t2*  2.080898e-05 -3.481513e-08 4.916633e-06
## t3*  5.647103e-03  1.665079e-05 2.318937e-04

(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.

• The bootstrap standard error seems to be about half of the glm summary which is great news!

Problem 9

We will now consider the Boston housing data set, from the ISLR2 library.

library(MASS)

## 
## Attaching package: 'MASS'

## The following object is masked from 'package:ISLR2':
## 
##     Boston

detach(Default)
attach(Boston)

(a) Based on this data set, provide an estimate for the population mean of medv. Call this estimate ˆµ(=mu).

mu=mean(medv)
mu

## [1] 22.53281

(b) Provide an estimate of the standard error of ˆµ. Interpret this result. Hint: We can compute the standard error of the sample mean by dividing the sample standard deviation by the square root of the number of observations.

sd(medv)/sqrt(length(medv))

## [1] 0.4088611

• The standard error of mu is estimated at 40.88%

(c) Now estimate the standard error of ˆµ using the bootstrap. How does this compare to your answer from (b)?

boot.fn=function(data,index){
  mean(data[index])
}
boot(medv,boot.fn,1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 -0.01309763   0.4050936

• Our bootstrap standard error is 40.5% which is a slight increase from the 40.88% we received from the original standard error. Although, it is not any significant improvement.

(d) Based on your bootstrap estimate from (c), provide a 95 % confidence interval for the mean of medv. Compare it to the results obtained using t.test(Boston$medv). Hint: You can approximate a 95 % confidence interval using the formula [ˆµ − 2SE(ˆµ), µˆ + 2SE(ˆµ)].

t.test(Boston$medv)

## 
##  One Sample t-test
## 
## data:  Boston$medv
## t = 55.111, df = 505, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  21.72953 23.33608
## sample estimates:
## mean of x 
##  22.53281

mu

## [1] 22.53281

se=0.4088
mu-2*se

## [1] 21.71521

mu+2*se

## [1] 23.35041

• These are very similar, arguably the same.

(e) Based on this data set, provide an estimate, ˆµmed, for the median value of medv in the population.

median(medv)

## [1] 21.2

(f) We now would like to estimate the standard error of ˆµmed. Unfortunately, there is no simple formula for computing the standard error of the median. Instead, estimate the standard error of the median using the bootstrap. Comment on your findings.

boot.fn=function(data,index){
  median(data[index])
}
boot(medv,boot.fn,1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 -0.0062   0.3713886

• The standard error of the median is actually lower than that of the mean.

(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston census tracts. Call this quantity ˆµ0.1. (You can use the quantile() function.)

quantile(medv,0.1)

##   10% 
## 12.75

(h) Use the bootstrap to estimate the standard error of ˆµ0.1. Comment on your findings.

boot.fn=function(data,index){
  quantile(data[index],0.1)
}
boot(medv,boot.fn,1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75 0.00135   0.5001658

• This is the highest standard error rate so far at 50! Much higher than the bootstrap errors for median and mean.